OFFSET
1,1
COMMENTS
This sequence is equivalent to all of the following sets (written in increasing order):
- all integers the form (p^e)(k^2) for p prime congruent to 3 (mod 8), e congruent to 1 (mod 4), and k an odd number coprime to p;
- all integers with an odd number of representations as x^2 + 2y^2 for odd x and y; and
- elements of A192628 which are congruent to 3 (mod 8).
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000
Joshua N. Cooper, Dennis Eichhorn and Kevin O'Bryant, Reciprocals of binary power series, International Journal of Number Theory, Vol. 2, No. 4 (2006), pp. 499-522; arXiv preprint, arXiv:math/0506496 [math.NT], 2005.
Joshua N. Cooper and Alexander W. N. Riasanovsky, On the Reciprocal of the Binary Generating Function for the Sum of Divisors, J. Int. Seq., Vol. 16 (2013), Article #13.1.8; alternative link.
EXAMPLE
3 is in the sequence since 3 = (3^1)(1^2); 3 is prime and congruent to 3 (mod 8), 1 is congruent to 1 (mod 4), and 1 is an odd integer coprime to 3.
6 is not in the sequence: since it is squarefree, k must be 1, but 6 cannot be written as p^e.
27 is not in the sequence: the only possible values for k are 1 and 3. In the k=1 case, 27 = (3^3)(1^2) does not work since e = 3 is not congruent to 1 (mod 4), and in the k=3 case, 27 = (3^1)(3^2), k=3 and p=3 are not coprime.
243 is in the sequence since 243 = (3^5)(1^2); 3 is prime and congruent to 3 (mod 8), 5 is congruent to 1 (mod 4), and 1 is an odd integer coprime to 3.
MATHEMATICA
ofTheFormQ[n_] := If[Length[fin = FactorInteger[n]] == 1 && Mod[fin[[1, 1]], 8] == 3 && Mod[fin[[1, 2]], 4] == 1, True, pe = Times @@ Power @@@ ({#[[1]], Mod[#[[2]], 2]} & /@ fin); k = Sqrt[n/pe]; fip = FactorInteger[pe]; Length[fip] == 1 && Mod[p = fip[[1, 1]], 8] == 3 && Mod[e = fip[[1, 2]], 4] == 1 && OddQ[k] && CoprimeQ[k, p]]; Select[Range[1, 999, 2], ofTheFormQ] (* Jean-François Alcover, Jan 22 2013 *)
PROG
(Sage)
prec = 2^10
L = []
for n in range(1, prec, 2):
n = Integer(n)
sfp = n.squarefree_part()
if mod(sfp, 8) == 3 and sfp.is_prime() and mod(n.ord(sfp), 4) == 1:
L.append(n)
print(L)
(Sage)
def BPS(n): #binary power series
return sum([q^s for s in n])
prec = 2^14
R = PowerSeriesRing(GF(2), 'q', default_prec = prec)
q = R.gen()
dList = [(2*n+1)^2 for n in range(0, (sqrt(prec)-1)/2)]
dSeries = BPS(dList)
print((dSeries^3).exponents()[:128])
CROSSREFS
KEYWORD
nonn
AUTHOR
Alexander Riasanovsky, Dec 31 2012
STATUS
approved